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Sloan [31]
2 years ago
10

Grade 10 Math. Solve for y. Will mark right answer brainliest :)​

Mathematics
1 answer:
Dima020 [189]2 years ago
5 0

Answer:

y=5, y=\frac{38}{11}

Step-by-step explanation:

Hi there!

We are given the equation

\frac{y+2}{y-3}+\frac{y-1}{y-4}=\frac{15}{2} and we need to solve for y

first, we need to find the domain, which is which is the set of values that y CANNOT be, as the denominator of the fractions cannot be 0

which means that y-3≠0, or y≠3, and y-4≠0, or y≠4

\frac{y+2}{y-3} and \frac{y-1}{y-4} are algebraic fractions, meaning that they are fractions (notice the fraction bar), but BOTH the numerator and denominator have algebraic expressions

Nonetheless, they are still fractions, and we need to add them.

To add fractions, we need to find a common denominator

One of the easiest ways to find a common denominator is to multiply the denominators of the fractions together

Let's do that here;

on \frac{y+2}{y-3}, multiply the numerator and denominator by y-4

\frac{(y+2)(y-4)}{(y-3)(y-4)}; simplify by multiplying the binomials together using FOIL to get:

\frac{y^{2}-2y-8}{y^{2}-7y+12}

Now on \frac{y-1}{y-4}, multiply the numerator and denominator by y-3

\frac{(y-1)(y-3)}{(y-4)(y-3)}; simplify by multiplying the binomials together using FOIL to get:

\frac{y^{2}-4y+3}{y^{2}-7y+12}

now add \frac{y^{2}-2y-8}{y^{2}-7y+12} and \frac{y^{2}-4y+3}{y^{2}-7y+12} together

Remember: since they have the same denominator, we add the numerators together

\frac{y^{2}-2y-8+y^{2}-4y+3}{y^{2}-7y+12}

simplify by combining like terms

the result is:

\frac{2y^{2}-6y-5}{y^{2}-7y+12}

remember, that's set equal to \frac{15}{2}

here is our equation now:

\frac{2y^{2}-6y-5}{y^{2}-7y+12}=\frac{15}{2}

it is a proportion, so you may cross multiply

2(2y²-6y-5)=15(y²-7y+12)

do the distributive property

4y²-12y-10=15y²-105y+180

subtract 4y² from both sides

-12y-10=11y²-105y+180

add 12 y to both sides

-10=11y²-93y+180

add 10 to both sides

11y²-93y+190=0

now we have a quadratic equation

Let's solve this using the quadratic formula

Recall that the quadratic formula is y=(-b±√(b²-4ac))/2a, where a, b, and c are the coefficients of the numbers in a quadratic equation

in this case,

a=11

b=-93

c=190

substitute into the formula

y=(93±√(8649-4(11*190))/2*11

simplify the part under the radical

y=(93±√289)/22

take the square root of 289

y=(93±17)/22

split into 2 separate equations:

y=\frac{93+17}{22}

y=\frac{110}{22}

y=5

and:

y=\frac{93-17}{22}

y=\frac{76}{22}

y=\frac{38}{11}

Both numbers work in this case (remember: the domain is y≠3, y≠4)

So the answer is:

y=5, y=\frac{38}{11}

Hope this helps! :)

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Step-by-step explanation:

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5/1 = 5

Since reciprocal is asking you to flip 5/1 meaning numerator becomes denominator and vice versa you end up with 1/5 being the reciprocal of 5

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2 years ago
In Exercises 5-7, find all the exact t-values for which the given statement is true,
bearhunter [10]

Answer:  See Below

<u>Step-by-step explanation:</u>

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians:     t = 0π + 2πn

In degrees:   t = 0° + 360n

******************************************************************************

6)\quad sin (t) = \dfrac{1}{2}

Where on the Unit Circle does   sin = \dfrac{1}{2}

<em>Hint: sin is only positive in Quadrants I and II</em>

\text{Answer: at}\  \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)

\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n

In degrees:    t = 30° + 360n  and  150° + 360n

******************************************************************************

7)\quad tan (t) = -\sqrt3

Where on the Unit Circle does    \dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)

<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>

\text{Answer: at}\  \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)

\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n

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4 0
2 years ago
Helpp me with this please!
Dmitry [639]
It’s CED hope this helps
6 0
3 years ago
in wereda election where four candidates appeared for election, the winning candidate received 36,000 votes which represented 45
Dmitry [639]

Let's see

Total votes be x

  • 45% of x=3600
  • 0.45x=3600
  • x=3600/0.45
  • x=8000

Now total percentage of vote

  • 45+20+25+6=96%

Not voted=4%

Find not voted

  • 0.04(8000)
  • 320

Total voted

  • 8000-320
  • 7680
7 0
1 year ago
Read 2 more answers
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